Conserved Quantity Along Affine Parameter

In summary, we discussed the Lagrangian in usual Schwarzschild coordinates and found that the coordinates for t, ϕ, and λ do not appear in the Lagrangian, leading to three conserved quantities: E, L, and Q. While E and L are physically meaningful as conserved energy and angular momentum, Q is not as it only serves to make λ an affine parameter. Different values of Q correspond to different scalings for λ and can be arbitrarily chosen. By inserting the other constants of motion into Q, we can derive the equation of motion for r, which is similar to the classical Newtonian equation with some additional terms.
  • #1
35,960
14,450
TL;DR Summary
What is the meaning/interpretation of the conserved quantity along the affine parameter in the Schwarzschild spacetime
In the usual Schwarzschild coordinates the Lagrangian can be written: $$\mathcal{L}= \frac{\dot r^2}{1-\frac{2M}{r}} - \left( 1- \frac{2M}{r} \right) \dot t^2 + r^2 \dot \phi^2$$ where all derivatives are with respect to a (affine) parameter ##\lambda##, and where for convenience I have considered units such that ##c=G=1## and coordinates such that ##\theta = \pi/2## so everything is in the equatorial plane.

Inspecting the Lagrangian we see that ##t##, ##\phi##, and ##\lambda## do not appear. So we have three easy conserved quantities: $$ E=\left( 1-\frac{2M}{r} \right) \dot t^2$$ $$ L=r^2 \dot \phi $$ $$Q = \frac{\dot r^2}{1-\frac{2M}{r}}-\left( 1-\frac{2M}{r} \right) \dot t^2 + r^2 \dot \phi^2$$

I understand that ##E## is interpreted as a conserved energy and ##L## is interpreted as a conserved angular momentum. But what is ##Q##?

It is conserved, but it isn't apparent to me what it is. I also am not sure if it is useful. I can solve for ##\dot t## in terms of ##E## and for ##\dot \phi## in terms of ##L## and use those to simplify my geodesics. But I don't see anything similar to be done for ##Q##.
 
Physics news on Phys.org
  • #2
Conservation of the squared 4-velocity. Essentially the affine parameter requirement.
 
  • #3
Oh, so it just is the condition in these coordinates that makes ##\lambda## into an affine parameter instead of some generic parameter.

I guess then that different values for ##Q## correspond to different scalings for the affine parameter. So the value of ##Q## can be arbitrarily and freely selected and is not physically meaningful like ##E## or ##L##.
 
  • #4
Dale said:
I guess then that different values for ##Q## correspond to different scalings for the affine parameter. So the value of ##Q## can be arbitrarily and freely selected and is not physically meaningful like ##E## or ##L##.
The standard choice would of course be ##Q =-1## or 0 (the latter for light-like geodesics). Note that ##Q = \mathcal L##.

Inserting the other constants of motion into ##Q## gives the typical effective equation of motion for ##r## which is very reminiscent of the classical Newtonian one (with some additions).
 
  • Like
Likes Dale and malawi_glenn
  • #5
Dale said:
But what is ?
##Q## is just the Lagrangian itself. Would we ever expect the parameter ##\lambda## to appear in the Lagrangian?
 
  • Like
Likes Dale and malawi_glenn
  • #6
Orodruin said:
Note that ##Q = \mathcal L##
PeterDonis said:
##Q## is just the Lagrangian itself
Embarrassingly, I didn't even notice that.

PeterDonis said:
Would we ever expect the parameter ##\lambda## to appear in the Lagrangian?
No. This is part of what confused me. I can easily think of Lagrangians without conserved energy or angular momentum, but not ##Q##. I guess it is unsurprising that I could not think of a Lagrangian without a Lagrangian. :doh:
 
  • Like
Likes malawi_glenn
  • #7
Orodruin said:
Inserting the other constants of motion into ##Q## gives the typical effective equation of motion for ##r## which is very reminiscent of the classical Newtonian one (with some additions).
Interesting. So I get $$ \frac{r^3 \left(\dot r^2-E^2\right)-2 L^2 M+L^2 r}{r^2 (r-2 M)}=-1 $$ I could solve that for ##\dot r## to get a first-order differential equation that seems useful. Otherwise, even with the conserved quantities, the original Euler equations gives me a second-order differential equation for ##r##.
 
  • #8
This is from my GR lecture notes:
1658328140871.png

In essence, you find the Newtonian equation of motion with time replaced by proper time and the additional term caused by ##\alpha## (your ##Q##, essentially a constant term) and the additional term caused by the cross term of the angular momentum barrier with the Newtonian potential.

Edit: Note that I also defined ##E## differently ...
1658328341267.png

This choice is obviously just to reproduce the similarity to the Newtonian equation.

Edit 2: You should find the same if you replace ##E^2 \to 2E## and solve for ##\dot r^2/2##.
 
  • Informative
  • Like
Likes vanhees71 and Dale
  • #9
Dale said:
Oh, so it just is the condition in these coordinates that makes ##\lambda## into an affine parameter instead of some generic parameter.

I guess then that different values for ##Q## correspond to different scalings for the affine parameter. So the value of ##Q## can be arbitrarily and freely selected and is not physically meaningful like ##E## or ##L##.
Yes, that's the great thing when choosing the "squared form" of the Lagrangian. Your parameter is automatically affine since ##Q=\text{const}## means that ##g_{\mu \nu} \dot{q}^{\mu} \dot{a}^{\nu}=\text{const}##. If you have massive particle, you can choose this to be ##c^2##, and your affine parameter is the proper time of the particle. You can of course also choose ##Q=0## for a massless particle.
 
  • Like
Likes Dale

Similar threads

Replies
1
Views
623
Replies
10
Views
2K
Replies
11
Views
1K
Replies
4
Views
2K
Replies
2
Views
1K
Replies
3
Views
2K
Replies
28
Views
3K
Replies
5
Views
2K
Back
Top